Answer:
Time take to fill the standing wave to the entire length of the string is 1.3 sec.
Explanation:
Given :
The length of the one end 
, frequency of the wave 
= 2.3 Hz, wavelength of the wave λ = 1 m.
Standing wave is the example of the transverse wave, standing wave doesn't transfer energy in a medium.
We know,
∴ 
λ
Where 
speed of the standing wave.
also, ∴ 
where 
time take to fill entire length of the string.
Compare above both equation,
⇒ 
sec
Therefore, the time taken to fill entire length 0f the string is 1.3 sec.
Answer:
A. If the sum of the external forces on an object is zero, then the object must be in equilibrium
Explanation:
Equilibrium, in physics, the condition of a system when neither its state of motion nor its internal energy state tends to change with time.
For a single particle, equilibrium arises if the vector sum of all forces acting upon the particle is zero.
the object is at equilibrium, then the net force acting upon the object should be 0 Newton. Thus, if all the forces are added together as vectors, then the resultant force (the vector sum) should be 0 Newton.
There are three types of equilibrium: stable, unstable, and neutral
Answer:
Explanation:
a. The equation of Lorentz transformations is given by:
x = γ(x' + ut')
x' and t' are the position and time in the moving system of reference, and u is the speed of the space ship. x is related to the observer reference.
x' = 0
t' = 5.00 s
u =0.800 c,
c is the speed of light = 3×10⁸ m/s
Then,
γ = 1 / √ (1 - (u/c)²)
γ = 1 / √ (1 - (0.8c/c)²)
γ = 1 / √ (1 - (0.8)²)
γ = 1 / √ (1 - 0.64)
γ = 1 / √0.36
γ = 1 / 0.6
γ = 1.67
Therefore, x = γ(x' + ut')
x = 1.67(0 + 0.8c×5)
x = 1.67 × (0+4c)
x = 1.67 × 4c
x = 1.67 × 4 × 3×10⁸
x = 2.004 × 10^9 m
x ≈ 2 × 10^9 m
Now, to find t we apply the same analysis:
but as x'=0 we just have:
t = γ(t' + ux'/c²)
t = γ•t'
t = 1.67 × 5
t = 8.35 seconds
b. Mavis reads 5 s on her watch which is the proper time.
Stanley measured the events at a time interval longer than ∆to by γ,
such that
∆t = γ ∆to = (5/3)(5) = 25/3 = 8.3 sec which is the same as part (b)
c. According to Stanley,
dist = u ∆t = 0.8c (8.3) = 2 x 10^9 m
which is the same as in part (a)
